Help with my Physic HW?

As shown in the figure , a superball with mass m equal to 50 grams is dropped from a height of h_{\rm i} = 1.5 \;{\rm m} . It collides with a table, then bounces up to a height of h_{\rm f}= 1.0\; {\rm m} . The duration of the collision (the time during which the superball is in contact with the table) is t_{\rm c} = 15 \;{\rm ms} . In this problem, take the positive y direction to be upward, and use g = 9.8\; {\rm m/s^2} for the magnitude of the acceleration due to gravity. Neglect air resistance.

A. Find the y component of the momentum, p_{{\rm before}{,}y}, of the ball immediately before the collision.

B.Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.

C.Find the y component of the time-averaged force F_{{\rm avg}{,}y}, in newtons, that the table exerts on the ball.

D. Find J_y, the y component of the impulse imparted to the ball during the collision.

E. Find K_{\rm after}-K_{\rm before}, the change in the kinetic energy of the ball during the collision, in joules.

Any help would be appreciated.

Answer:

A. Find the y component of the momentum, p_{{\rm before}{,}y}, of the ball immediately before the collision.

= 0.05*sq rt(2*9.8*1.5) = 0.27 kg m s^-1

B.Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table. = 0.05*sq rt(2*9.8*1.0) = 0.22 kg m s^-1

C.Find the y component of the time-averaged force F_{{\rm avg}{,}y}, in newtons, that the table exerts on the ball.= [0.22 -(- 0.27)]/0.015 = 32.7 N

D. J_y, the y component of the impulse imparted to the ball during the collision=0.22+0.27=0.49 N

E. Find K_{\rm after}-K_{\rm before}, the change in the kinetic energy of the ball during the collision, in joules. (0.27^2 – 0.22^2)/(2*0.05) = 0.25 J